我有一张桌子
create table us
(
a number
);
现在我有如下数据:
a
1
2
3
4
null
null
null
8
9
现在我需要一个查询来计算列a中的空值和非空值
我有一张桌子
create table us
(
a number
);
现在我有如下数据:
a
1
2
3
4
null
null
null
8
9
现在我需要一个查询来计算列a中的空值和非空值
当前回答
对于非空值
select count(a)
from us
null值
select count(*)
from us
minus
select count(a)
from us
因此
SELECT COUNT(A) NOT_NULLS
FROM US
UNION
SELECT COUNT(*) - COUNT(A) NULLS
FROM US
应该做这项工作
更好的是列标题是正确的。
SELECT COUNT(A) NOT_NULL, COUNT(*) - COUNT(A) NULLS
FROM US
在我的系统上进行的一些测试中,需要进行全表扫描。
其他回答
为了提供另一种选择,Postgres 9.4+允许对聚合应用FILTER:
SELECT
COUNT(*) FILTER (WHERE a IS NULL) count_nulls,
COUNT(*) FILTER (WHERE a IS NOT NULL) count_not_nulls
FROM us;
SQLFiddle: http://sqlfiddle.com/ # !17/80a24/5
这里有两种解决方案:
Select count(columnname) as countofNotNulls, count(isnull(columnname,1))-count(columnname) AS Countofnulls from table name
OR
Select count(columnname) as countofNotNulls, count(*)-count(columnname) AS Countofnulls from table name
SELECT SUM(NULLs) AS 'NULLS', SUM(NOTNULLs) AS 'NOTNULLs' FROM
(select count(*) AS 'NULLs', 0 as 'NOTNULLs' FROM us WHERE a is null
UNION select 0 as 'NULLs', count(*) AS 'NOTNULLs' FROM us WHERE a is not null) AS x
这很糟糕,但它将返回一个带有2个cols的记录,指示null和非null的计数。
在阿尔贝托的基础上,我添加了汇总。
SELECT [Narrative] = CASE
WHEN [Narrative] IS NULL THEN 'count_total' ELSE [Narrative] END
,[Count]=SUM([Count]) FROM (SELECT COUNT(*) [Count], 'count_nulls' AS [Narrative]
FROM [CrmDW].[CRM].[User]
WHERE [EmployeeID] IS NULL
UNION
SELECT COUNT(*), 'count_not_nulls ' AS narrative
FROM [CrmDW].[CRM].[User]
WHERE [EmployeeID] IS NOT NULL) S
GROUP BY [Narrative] WITH CUBE;
下面是一个在Oracle上运行的快速而简单的版本:
select sum(case a when null then 1 else 0) "Null values",
sum(case a when null then 0 else 1) "Non-null values"
from us